【BZOJ 4817】[SDOI2017] 树点涂色

相关链接

题目传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=4817

解题报告

我们发现涂色可以看作LCT的access操作
然后权值就是到根的虚边数

于是用LCT来维护颜色
用线段树配合DFS序来支持查询
时间复杂度:$O(n \log^2 n)$

Code

#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 100009;
const int M = N << 1;
const int LOG = 20;

int n, m, head[N], nxt[M], to[M]; 
int in[N], ot[N], dep[N], num[N], ff[N][LOG];

inline int read() {
	char c = getchar(); int ret = 0, f = 1;
	for (; c < '0' || c > '9'; f = c == '-'? -1: 1, c = getchar());
	for (; '0' <= c && c <= '9'; ret = ret * 10 + c - '0', c = getchar());
	return ret * f;
}

inline void AddEdge(int u, int v) {
	static int E = 1;
	to[++E] = v; nxt[E] = head[u]; head[u] = E;
	to[++E] = u; nxt[E] = head[v]; head[v] = E;
}

inline int LCA(int u, int v) {
	if (dep[u] < dep[v]) {
		swap(u, v);
	}
	for (int j = LOG - 1; ~j; j--) {
		if (dep[ff[u][j]] >= dep[v]) {
			u = ff[u][j];
		}
	}
	if (u == v) {
		return u;
	}
	for (int j = LOG - 1; ~j; j--) {
		if (ff[u][j] != ff[v][j]) {
			u = ff[u][j];
			v = ff[v][j];
		}
	}
	return ff[u][0];
}

class SegmentTree{
	int root, ch[M][2], tag[M], mx[M];
public:
	inline void init() {
		build(root, 1, n);
	}
	inline void modify(int l, int r, int d) {
		modify(root, 1, n, l, r, d);
	}
	inline int query(int l, int r = -1) {
		return query(root, 1, n, l, r >= 0? r: l);
	}
private:
	inline void PushDown(int w) {
		if (tag[w]) {
			int ls = ch[w][0], rs = ch[w][1];
			mx[ls] += tag[w];
			mx[rs] += tag[w];
			tag[ls] += tag[w];
			tag[rs] += tag[w];
			tag[w] = 0;
		}
	}
	inline int query(int w, int l, int r, int L, int R) {
		if (L <= l && r <= R) {
			return mx[w];
		} else {
			PushDown(w);
			int mid = l + r + 1 >> 1, ret = 0;
			if (L < mid) {
				ret = max(ret, query(ch[w][0], l, mid - 1, L, R));
			}
			if (mid <= R) {
				ret = max(ret, query(ch[w][1], mid, r, L, R));
			}
			return ret;
		}
	}
	inline void modify(int w, int l, int r, int L, int R, int d) {
		if (L <= l && r <= R) {
			tag[w] += d;
			mx[w] += d;
		} else {
			PushDown(w);
			int mid = l + r + 1 >> 1;
			if (L < mid) {
				modify(ch[w][0], l, mid - 1, L, R, d);
			}
			if (mid <= R) {
				modify(ch[w][1], mid, r, L, R, d);
			}
			mx[w] = max(mx[ch[w][0]], mx[ch[w][1]]);
		}
	}
	inline void build(int &w, int l, int r) {
		static int cnt = 0;
		w = ++cnt;
		if (l == r) {
			mx[w] = dep[num[l]];
		} else {
			int mid = l + r + 1 >> 1;
			build(ch[w][0], l, mid - 1);
			build(ch[w][1], mid, r);
			mx[w] = max(mx[ch[w][0]], mx[ch[w][1]]);
		}
	}
}SGT;

class LinkCutTree{
	int ch[N][2], fa[N];
public:
	inline void SetFather(int w, int f) {
		fa[w] = f;
	}
	inline void access(int x) {
		for (int last = 0; x; last = x, x = fa[x]) {
			splay(x);
			if (fa[x]) {
				int p = GetMin(x);
				SGT.modify(in[p], ot[p], -1);
			}
			if (ch[x][1]) {
				int p = GetMin(ch[x][1]);
				SGT.modify(in[p], ot[p], 1);
			}
			ch[x][1] = last;
		}
	}
private:
	inline bool IsRoot(int x) {
		return !fa[x] || (ch[fa[x]][0] != x && ch[fa[x]][1] != x);
	}
	inline int GetMin(int x) {
		return ch[x][0]? GetMin(ch[x][0]): x;
	}
	inline void splay(int x) {
		for (int f, ff; !IsRoot(x); ) {
			f = fa[x], ff = fa[f];
			if (IsRoot(f)) {
				rotate(x);
			} else {
				if ((ch[ff][0] == f) ^ (ch[f][0] == x)) {
					rotate(x);
					rotate(x);
				} else {
					rotate(f);
					rotate(x);
				}
			}
		}
	}
	inline void rotate(int x) {
		int f = fa[x], t = ch[f][1] == x;
		fa[x] = fa[f];
		if (!IsRoot(f)) {
			ch[fa[f]][ch[fa[f]][1] == f] = x;
		}
		ch[f][t] = ch[x][t ^ 1];
		fa[ch[x][t ^ 1]] = f;
		ch[x][t ^ 1] = f;
		fa[f] = x;
	}
}LCT;

inline void DFS(int w, int f) {
	static int ID = 0;
	LCT.SetFather(w, f);
	ff[w][0] = f;
	dep[w] = dep[f] + 1;
	num[in[w] = ++ID] = w;
	for (int i = head[w]; i; i = nxt[i]) {
		if (to[i] != f) {
			DFS(to[i], w);
		}
	}
	ot[w] = ID;
}	

int main() {
	n = read(); m = read();
	for (int i = 1; i < n; i++) {
		AddEdge(read(), read());
	}
	DFS(1, 0);
	for (int j = 1; j < LOG; j++) {
		for (int i = 1; i <= n; i++) {
			ff[i][j] = ff[ff[i][j - 1]][j - 1];
		}
	}
	SGT.init();
	for (int i = 1; i <= m; i++) {
		int opt = read();
		if (opt == 1) {
			LCT.access(read());
		} else if (opt == 2) {
			int u = read(), v = read(), lca = LCA(u, v);
			printf("%d\n", SGT.query(in[u]) + SGT.query(in[v]) - 2 * SGT.query(in[lca]) + 1);
		} else {
			int x = read();
			printf("%d\n", SGT.query(in[x], ot[x]));
		}
	}
	return 0;
}

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